The Cayley-Hamilton Theorem

For an $n \times n$ matrix $A$, we have already seen one polynomial: the characteristic polynomial $\chi_A$.

There is another important polynomial associated to $A$: its minimal polynomial $m_A$.

Definition. Let $A$ be a $n \times n$ matrix over a field $k$ and let $p(x)$ be a polynomial with coefficient in $k$. We say that $A$ satisfies $p(x)$ if \(p(A) = 0\)

Let’s look at some examples.

Example.

• Let’s take $A = I_n$ the identity matrix. Then, $A$ satisfies the polynomial $x-1$. Notice that to make sense of plugging in a matrix we need to promote $1$ to a $n \times n$ matrix itself. We do this by replace the constant term $c$ by $cI_n$. Thus, \(I_n - I_n = 0\)

• Let’s take \(A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\) Then, $A$ will not satisfy any polynomial of degree $1$ since it is not equal to $cI_2$ for any scalar $c$. Let’s try to find a degree polynomial that $A$ satisfies. We have \(A^2 = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}\) Note that \(A^2 -2A + 2I_2 = 0\) Thus $A$ satisfies $x^2 - 2x + 2$. You might notice that this is also the characteristic polynomial of $A$.

Theorem (Cayley-Hamilton). A $n \times n$ matrix satisfies its own characteristic polynomial.

Unforunately, we don’t have time to delve into a proof of this result but it involves some fundamental ideas in modern algebra.

We will content ourselves with observing some consequences. A polynomial is called monic if its leading coefficient is $1$, ie \(p(x) = x^d + a_{d-1} x^{d-1} + \cdots + a_0\)

Definition. The minimal polynomial of $A$ is the minimal degree monic polynomial that $A$ satisfies. We denote as $m_A(x)$.

In both our examples, we have actually written down minimal polynomials.

Corollary. The degree of $m_A(x)$ is at most $n$.

In fact, one can deduce that there is another polynomial $r(x)$ such that \(\chi_A(x) = m_A(x) r(x)\) so that $m_A(x)$ actually divides $\chi_A(x)$.

We also know how the roots of the two polynomials compare.

Theorem. The roots of $m_A(x)$ and the roots $\chi_A(x)$ are equal sets.

Over $\mathbb{C}$, due to the Fundamental Theorem of Algebra we know that \(\chi_A(x) = \prod_{i=1}^s (x-\alpha_i)^{e_i}\) for scalars $\alpha_i \in \mathbb{C}$ and integers $e_i \geq 1$. Then, the previous theorem says, \(m_A(x) = \prod_{i=1}^s (x-\alpha_i)^{d_i}\) where $1 \leq d_i \leq e_i$ for each $i$.

Let’s see how these compare in some examples.

Example.

• For the identity matrix $I_n$, we have \(\chi_{I_n} = (1-x)^n\) and \(m_{I_n}(x) = x-1\) so clearly $m_{I_n} \mid \chi_{I_n}$ and they are pretty far from equal.

• Consider the matrix \(A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\) Then, $A$ has the same characteristic polynomial as $I_2$ \(\chi_A(x) = (1-x)^2\) but its minimal polynomial is \(m_A(x) = (1-x)^2 = (x-1)^2\) equal to its characteristic polynomial.